%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[12pt,a4paper]{article} %-- used packages ------------------------------------------------------ %%%JEC 3/2/06 does not compile at LAL \usepackage{cite} %\usepackage[T1]{fontenc} %\usepackage[latin1]{inputenc} \usepackage{graphicx} \usepackage{epsfig} \usepackage{amssymb} \usepackage{amsmath} \usepackage{latexsym} \setlength{\textheight}{23cm} \setlength{\textwidth}{16cm} \setlength{\topmargin}{0cm} \setlength{\oddsidemargin}{0cm} \setlength{\evensidemargin}{0cm} \setlength{\marginparwidth}{0pt} \def\beq{\begin{equation}} \def\eeq{\end{equation}} \def\bea{\begin{eqnarray}} \def\eea{\end{eqnarray}} \begin{document} \title{Dark Current induced trigger in MEMPHYS\thanks{Preliminary}} \author{J.E Campagne - LAL - Orsay} \date{\today} \maketitle \begin{abstract} Blabla \end{abstract} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Motivations} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% On 5th of July, J.~Bouchez had demonstrated the need to investigate deeper than expected the dark current induced trigger for a MEMPHYS tank. As a matter of fact, a 3~MeV reactor neutrino produce around 15~p.e, so basically 15 PMTs on a Cerenkov cone are hit, while in the same time at 2~kHz dark current noise per PMT in a 400~ns time window one can expect around 160 PMT hit randomly distributed. Although, it may be not a problem (not yet proved) to recover the PMTs of the signal, the main concern is the trigger. J.~Bouchez has made a preliminary numerical analysis to show that the question is relevant. A long this line, I have investigated a toy MC and analytical analysis. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{A toy MC} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The toy MC consists simply of the generation of dark current random noise on each PMT baseline, then it simulates a digital sum of all the PMT baselines, and applies a threshold on a total number of PMT hit during a sliding window. This toy MC can be adapted according to: \begin{verbatim} NumberOfPMT = 81,000 for MEMPHYS DarkCurrentRate = 6.0*kHz more likely for 12" TriggerSlidingGate = 425.0*ns for MEMPHYS PulseWidth = 5.0*ns this is the time quantum (a pulse width equivalent) TimeDurationOfExp = 6000.0*micros (+TriggerSlidingGate) as a total time for investigation CountThreshold = number of PMTs at least to be fired in a TriggerSlidingGate to produce a Trigger \end{verbatim} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Dark current generation} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% According to the fact that dark current is a random noise, if ones make the assumption that there is no correlation between two noise hit (at least on the scale of \verb|TriggerSlidingGate|), then the time interval between two hit $\delta t$ follow an exponential probability with parameter $\tau = 1/\verb|DarkCurrentRate|$. So, the probability to get 1 hit in the time window $[t_0,t_0+\Delta t]$ reads: \bea P(1;[t_0,t_0+\Delta t]) & = & \int_{t_0}^{t_0+\Delta t} e^{(t-t_0)/\tau} \frac{dt}{\tau} \nonumber \\ & = & 1 - e^{-\Delta t/\tau} = \frac{\Delta t}{\tau} - \frac{1}{2} \left( \frac{\Delta t}{\tau}\right)^2 \dots \label{eq:singleProba} \eea and is independent of $t_0$. If one piles up $N$ PMTs and if $\Delta t/\tau \rightarrow 0$, then it is well known that the probability to get $n$ PMTs hit in a time window of $\Delta t$ tends to a Poisson distribution: \beq P(n|N;[t,t+\Delta t]) \stackrel{\Delta t/\tau \rightarrow 0}{\longrightarrow}\mathrm{Poisson}(n;N\Delta t/\tau) \label{eq:poissonLimit} \eeq One can see these properties reproduced by the toy MC on Fig.~\ref{fig:PMTPoisson} for some parameters values used on purpose. \begin{figure}[htb] \begin{minipage}[c]{0.44\textwidth} \includegraphics[width=\textwidth]{deltaT.eps} \end{minipage} \begin{minipage}[c]{0.05\textwidth} \end{minipage} \begin{minipage}[c]{0.44\textwidth} \includegraphics[width=\textwidth]{NumberPMTs.eps} \end{minipage} \caption{\label{fig:PMTPoisson} {\it left panel}: exponential generation of $\delta t$ with $1/\tau = 1$~MHz dark current noise; {\it right panel}: number of PMTs fired among 50 PMTs, per $\Delta t = 100$~ns windows. One expects a Poisson distribution with a mean of $N\Delta t/\tau = 5$ as it is shown. } \end{figure} In fact, as it is mentioned, the Poisson distribution is an asymptotic distribution. The real probability to get exactly $n$ PMTs fired among a collection of $N$ PMTs in a $\Delta t$ time window, with the hypothesis that each PMT dark current rate has a $1/\tau$ frequency, reads (using Eq.~\ref{eq:singleProba}): \bea P(n|N;[t,t+\Delta t]) & = & C_N^n P(1;[t,t+\Delta t])^n \left(1-P(1;[t,t+\Delta t])\right)^{N-n} \nonumber \\ & = &\mathrm{Binomial}(n,N;P(1,[t,t+\Delta t])) \label{eq:BinomialProb} \eea %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Trigger simulation} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% One may wish to know the occurrence of "how many times per second, $n_{Th}$ PMTs are fired among $N$ PMTs?". In fact, J.~Bouchez noticed that this question is not complete as one may trigger only if the number of PMTs fired in a $\Delta t$ sliding windows just exceeds $n_{Th}$. That is to say, one should consider the stepping function of the number of PMTs fired in the sliding window $\Delta t$ examined at $t$, noted $n(t)$, and consider as "trigger" the passage $n(t)=n_{Th} \rightarrow n(t+dt)=n_{Th}+1$ in an infinitesimal time duration $dt$. In fact, looking at Fig.~\ref{fig:DigitalSum}, where is displayed the digital sum of $N=81,000$ PMts with 6~kHz dark current noise rate in sliding window of $425$~ns (computed every $dt=5$~ns), one notices that it is true that the threshold\footnote{mean number of PMTS is 206 and the threshold is set 3 sigma above} at $250$ is reached by steps, but one may also specify that if there is a trigger at $t_1$, before to re-trigger the $n(t>t_1)$ function may pass through a sequence $n(t)=n_{Th}+1 \rightarrow n(t+dt)=n_{Th}$. \begin{figure}[htb] \begin{minipage}[c]{0.44\textwidth} \includegraphics[width=\textwidth]{SumBaseline.eps} \end{minipage} \begin{minipage}[c]{0.05\textwidth} \end{minipage} \begin{minipage}[c]{0.44\textwidth} \includegraphics[width=\textwidth]{SumBaselineZoom.eps} \end{minipage} \caption{\label{fig:DigitalSum} {\it left panel}: Digital sum of the time evolution of $N=81,000$ PMTs with a dark current of $6$~kHz, computed for a time sliding window of $425$~ns every $5$~ns; {\it right panel}: zoom around the location of a trigger threshold at $250$ PMTs. } \end{figure} So in practice, one considers as a trigger the passage $n(t)=n_{Th} \rightarrow n(t+dt)=n_{Th}+1$ and groups together the following $n(t)$ values contiguous in time for which $n(t)>=n_{Th}$. Using this definition, one gets the results for one MEMPHYS tank (Tab.\ref{tab:TriggerRate}): $N=81,000$ PMTs, a sliding window of $425$~ns which corresponds to the maximum time of flight of \v{C}erenkov light in the $\phi: 65~\mathrm{m} \times H:65~\mathrm{m}$ cylinder water volume ($v = 22,6~\mathrm{cm/ns}$). \begin{table} \begin{center} \begin{tabular}{cccc}\hline\hline DC per PMT (kHz) & Mean of PMTs fired & Threshold (at 3$\sigma$) & Trigger (kHz) & Th (kHz)\\ 1 & 34 & 52 & 412 & 57\\ 2 & 69 & 93 & 378 & 64.9\\ 3 & 103 & 133 & 198 & 2.8 10^3\\ 4 & 138 & 172 & 378 & 7.1\\ 5 & 172 & 211 & 268 & 3.1\\ 6 & 206 & 249 & 416 & 1.5\\ \hline DC per PMT (kHz) & Mean of PMTs fired & Threshold & Trigger \& rel. err (kHz) \\ 1 & 34 & 40 (1$\sigma$) & 30 811 (0.7\%) & 1475.5 \\ 1 & 34 & 46 (2$\sigma$) & 4580 (2\%) & 363.4\\ 1 & 34 & 52 (3$\sigma$) & 412 (7\%) & 41.2 \\ 1 & 34 & 57 (4$\sigma$) & 28 (12\%) & 3.9\\ \hline\hline \end{tabular} \caption{\label{tab:TriggerRate} Mean number of PMTs ($m$)fired among $N=81,000$ PMTs on a time sliding window of $\Delta t=425$~ns corresponding to the dark current (DC) rate ($f$) per PMT (this is $N\Delta t * f$). Then, setting a threshold to $m+3\sqrt m$, one reads the trigger rate (see text for the definition of a trigger). The relative errors is also reported as it is not negligeable.} \end{center} \end{table} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Analytical formula} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}